Conventionally, there is a case where two actuators are used to drive a single driven body such that, if one of the actuators fails, the driven body continues to be driven by the other actuator.
In this case, the drive amounts of the actuators are changed depending on whether the actuators are operating normally or abnormally. When both actuators are normally operating, each of the actuators drives the driven body by a drive amount that is half the total drive amount indicated by a command value. Thus, the driven body is driven by the total drive amount indicated by the command value. If one of the actuators fails, the other actuator drives the driven body by the total drive amount indicated by the command value.
To implement the control described above, according to Patent Document 1, the following method is carried out by using the configuration illustrated in FIG. 1.
(1) A drive amount Ef1 of a first motor (actuator) 61 and a drive amount Ef2 of a second motor (actuator) 63 are detected, and further, a total drive amount Ef of the first and second motors 61 and 63 is detected from a detection value of the drive amount of a driven body.
(2) A first multiplier 65a outputs a value Ed1 obtained by multiplying a command value (a position command value in FIG. 1) Ep by a gain G1 (=ΔEf1/ΔEf) to a first subtractor 67a, while a second multiplier 65b outputs a value Ed2 obtained by multiplying a command value Ep by a gain G2 (=ΔEf2/ΔEf) to a second subtractor 67b. The symbol Δ in this case means change an amount (time change rate) of Ef, Ef1, and Ef2.
(3) The first subtractor 67a outputs the difference between Ed1 and Ef1 to a first controller 69a, while the second subtractor 67b outputs the difference between Ed2 and Ef2 to a second controller 69b. 
(4) The first controller 69a drives the first motor 61 by the amount indicated by (Ed1−Ef1), while the second controller 69b drives the second motor 63 by the amount indicated by (Ed2−Ef2).
Thereafter, the steps (1) to (4) described above are repeated.
In this manner, when the motors 61 and 63 are normal, the motors 61 and 63 are driven by the same amount, so that Ef1 and Ef2 are equal. Therefore, Ef denotes the total of Ef1 and Ef2, so that G1 (=ΔEf1/ΔEf) and G2 (=ΔEf2/ΔEf) will both be ½. Hence, the driven body is driven to a position indicated by Ep by both the first and second motors 61 and 63.
Meanwhile, if the first motor 61 fails and stops operating, then ΔEf will have the same value as that of ΔEf2, so that G2 (=ΔEf2/ΔEf) will be 1. Hence, the driven body is driven to the position indicated by Ep by the second motor 63. The same applies if the second motor fails and stops operating.